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Question

# A radioactive material decays by simultaneous emission of two particles with half-lives 1620 yr and 810 yr respectively. The time in years after which one-fourth of material remains, is :

A
1080 yr
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B
2340 yr
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C
4860 yr
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D
3240 yr
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Solution

## The correct option is A 1080 yrSince, from Rutherford-Soddy law, the number of atoms left after half lives is given byN=N0(12)nwhere, N0 is the original number of atoms.The number of half lives, n=timeofdecayeffectivehalf−lifeRelation between effective disintegration constant (λ) and half life (T)λ=ln2T∴λ1+λ2=ln2T1+ln2T2Effective half life,1T=1T1+1T2=11620+18101T=1+21620⇒T=540yr∴n=T540∴N=N0(12)t/540⇒NN0=(12)2=(12)t/540⇒t540=2⇒t=2×540=1080yr

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